Assignment 3 solutions

Assignment 3 solutions - MIT OpenCourseWare...

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MIT OpenCourseWare http://ocw.mit.edu 18.085 Computational Science and Engineering I Fall 2008 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms .
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18.085 - Mathematical Methods for Engineers I Prof. Gilbert Strang Solutions - Problem Set 3 Section 1.6 1 1 1) T = 1 2 ⎦ ⎥ u T T u = u 1 u 2 1 1 u 1 1 2 u 2 u 1 u 2 = u 1 u 2 u 1 + 2 u 2 = u 1 2 u 1 u 2 + ( u 1 u 2 + 2 u 2 2 ) = u 1 2 2 u 1 u 2 + 2 u 2 2 = ( u 1 u 2 ) 2 + u 2 2 > 0 # u T T u is positive deFnite as it is the sum of 2 squares ± ± 1 1 0 2 1 1 3) A = ² 0 1 1 A T A = ² 1 2 1 1 0 1 1 1 2 Au = 0 A T Au = 0 A is a singular matrix as it has only 2 linearly independent columns. Since A does not have full rank, A T A will have a zero pivot ± 2 1 1 A T A = ² 1 2 1 1 1 2 ± 2 1 1 ³ 0 3 / 2 3 / 2 row 2 row 2 + 1 row 1 = ² 2 0 3 / 2 3 / 2 row 3 row 3 + 1 2 row 1 ± 2 1 1 = ² 0 3 / 2 3 / 2 0 0 0 row 3 row 3 + row 2 A T A is only semideFnite # Au = 0, A T Au = 0 The nullvector of A and A T A would be a constant vector ± ⎡ ± ± 1 1 0 c 0 ² 0 1 1 ⎣ ² c = ² 0 1 0 1 c 0 ± ⎡ ± ± 2 1 1 c 0 ² 1 2 1 ⎣ ² c = ² 0 1 1 2 c 0 1
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c u = ± c where c is a constant # c 1 b 6) K = b 4 ⎦ ⎥ u T Ku = ² u 1 u 2 1 b u 1 b 4 u 2 ² u 1 + bu 2 = u 1 u 2 bu 1 + 4 u 2 = u 1 2 + bu 1 u 2 + bu 1 u 2 + 4 u 2 2 = u 1 2 + 2 bu 1 u 2 + 4 u 2 2 = ( u 1 + bu 2 ) 2 + (4 b 2 ) u 2 2 > 0 4 b 2 > 0 ( b + 2)( b 2) < 0 ± ± ± ± ± ± ± ± ± ± ± ± ± ± ± ± ± ± ± ± ± ± ± 2 2 2 < b < 2 # For semidefnite case, the borderline value b is 2 and 2 u T Ku = ( u 1 + bu 2 ) 2 (only one square) 1 5 b = 5, K = 5 4 By Gaussian Elimination 1 5 5 4 1 5 = 0 21 row 2 row 2 5 × row 1 The pivots are 1 and 21 The matrix is indefnite b = 5 # 11) f ( x, y ) = 2 xy f f = 2 y = 2 x x y 2 f 2 f 2 f = 2 , = 0 , = 0 x y x 2 y 2 2
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0 2 Hessian matrix, H = 2 0 a b x x y 2 xy b c y ax + by = x y bx + cy = ax 2 + bxy + bxy + cy 2 = ax 2 + 2 bxy + cy 2 By comparing coefficients, a = 0, c = 0 b = 1 0 1 The symmetric matrix that produces f ( x, y ) = 2 xy is S = 1 0 det( S I ) = 0 1 det = 0 1 2 1 = 0 ( 1)( + 1) = 0 = 1 or 1 The eigenvalues for matrix S are 1 and 1 # 16) Since A is positive deFnite, A can be diagonalized to A = S S 1 where S = eigenvector = eigenvalues AA 1 = I ( S S 1 ) A 1 = I A 1 = S 1 S 1 The eigenvalues of A are
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Assignment 3 solutions - MIT OpenCourseWare...

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